PLEASE PLEASE do both PLease!!! THANKS!!! **Linear Algebra: Transformations**Let \( A = \begin{bmatrix}\frac{1}{5} & 0 & 0 \\0 & \frac{1}{5} & 0 \\0 & 0 & \frac{1}{5} \\\end{bmatrix}\), \( u = \begin{bmatrix}15 \\10 \\-25 \\\end{bmatrix}\), and \( v = \begin{bmatrix}g \\d \\h \\\end{bmatrix}\).Define \( T: \mathbb{R}^3 \rightarrow \mathbb{R}^3 \) by \( T(x) = Ax \).Find \( T(u) \) and \( T(v) \).**Calculation**\[ T(u) = \_\_\_\_\_\_\_\_\_\_\_\_\_ \](Simplify your answer. Use integers or fractions for any numbers in the expression.) **Determine if** \( v \) **is in the set spanned by the columns of** \( B \).\[B = \begin{bmatrix}3 & -3 & 2 \\9 & -7 & 9 \\6 & 0 & 15 \\9 & -5 & 16 \\\end{bmatrix}\]\[v = \begin{bmatrix}-12 \\-20 \\24 \\-4 \\\end{bmatrix}\]---**Choose the correct answer below and, if necessary, fill in the answer box(es) to complete your choice.**- \( \textcircled{A} \) Vector \( v \) is not in the set spanned by the columns of \( B \) because the columns of \( B \), \( b_1 \), \( b_2 \), and \( b_3 \) are linearly independent. - \( \textcircled{B} \) Vector \( v \) is in the set spanned by the columns of \( B \) because the columns of \( B \) span \(\mathbb{R}^4 \).- \( \textcircled{C} \) Vector \( v \) is not in the set spanned by the columns of \( B \) because the reduced echelon form of the matrix formed by writing \( B \) with a fourth column equal to \( v \) is \(\square\).- \( \textcircled{D} \) Vector \( v \) is in the set spanned by the columns of \( B \) because \(\square b_1 + \square b_2 + \square b_3 = v \).

Answered: Let A = 0 00 G|5 0 0 0 , u = 1|5 15 10…

Let A = 0 00 G|5 0 0 0 , u = 1|5 15 10 - 25 and v= g d h Define T: R³ R³ by T(x) = Ax. Find T(u) and T(v). →>>

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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